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Find the number of quadratic polynomials ax² + bx + c such that:

a) a, b, c are distinct.

b) a, b, c ε {1, 2, 3, …2008}

c) x + 1 divides ax² + bx + c

Reasoning->b=a+c b ranges from 3 to 2008 x1+x2=n has n-1C1 solns. Using th elogic we get 2+3+..2007 solutions. A total of 2015027 solutions.here,condition 1 comes into picture telling us that a,b,c are distinct. Therefore the even values of b ie.4,6,8,..2008 will have one solution less ie.from the given solution subtract 1003 thus giving us 20104024

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What is the point of the post if you have the answer? The answer you give is copied and pasted from the site Chandru gives. If you don't understand it, please indicate what you don't understand. –  Ross Millikan May 16 '11 at 14:13
    
@Ross Yes i have taken this question from the site,the problem is i am not satisfied wiith the logic,thats why i wanted a different approach to tackle this question,i also pasted the solution given there but still logic is vauge.the point is i am in search of simple and clear solution.i dont think i am wrong. –  prem shekhar May 16 '11 at 16:05
    
if you are copying something from a source, you should link to the source. If you are not satisfied with the proof, you should describe why you aren't satisfied and what would satisfy you. –  Qiaochu Yuan May 16 '11 at 16:33

3 Answers 3

up vote 3 down vote accepted

Ross's answer was my first thought, but as a slightly-different alternative, use the remainder theorem:

The remainder when a polynomial $P(x)$ is divided by $x-k$ is $P(k)$.

So if $x+1=x-(-1)$ divides $P(x)=ax^2+bx+c$ (leaving remainder $0$), $P(-1)=0$, which should give the same condition on $a$, $b$, and $c$ that would result from long division.

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as c=b-a when c=1 we will have 2007 pairs((2,1),(3,2)….) when c=2 we will have 2005 pairs when c-3 we will have 2004 pairs… . . . when c=2007 no. of pair is 1. –  prem shekhar May 16 '11 at 13:31
    
i have applied the above logic but still i am not getting the correct answer. –  prem shekhar May 16 '11 at 13:31
2  
@prem: when c=1, you cannot have b=2, a=1 as they are not distinct. –  Ross Millikan May 16 '11 at 13:36
1  
@Isaac: this does lead the same place. A good way to look at it. –  Ross Millikan May 16 '11 at 13:47
    
Reasoning->b=a+c b ranges from 3 to 2008 x1+x2=n has n-1C1 solns.Using th elogic we get 2+3+..2007 solutions. A total of 2015027 solutions.here,condition 1 comes into picture telling us that a,b,c are distinct.Therefore the even values of b ie.4,6,8,..2008 will have one solution less ie.from the given solution subtract 1003 thus giving us 20104024. –  prem shekhar May 16 '11 at 13:49

Your question and answers to your question can be found here. For the answer see: abhitsian's Comment in this link:

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In fact his answer is copied from there, including the typos. –  Ross Millikan May 16 '11 at 14:05
    
@Ross: We should inform/warn him. I did my job. It seems already Arturo had given him suggestions of not to post questions in an imperative mode and show his work. –  user9413 May 16 '11 at 14:06
    
Yes i have taken this question from the site,the problem is i am not satisfied wiith the logic,thats why i wanted a different approach to tackle this question,i also pasted the solution given there but still logic is vauge.the point is i am in search of simple and clear solution.i dont think i am wrong. –  prem shekhar May 16 '11 at 16:05

Hint: try polynomial long division to find a condition on $a,b,c$. Then it is a combinatoric question how many sets of that kind there are.

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i have solved it but i am not getting the correct answer.can you pplease elobrate your approch –  prem shekhar May 16 '11 at 13:29
    
@prem: if you have a solution, it would be nice to show what you have done. My answer (and Isaac's) may have been superfluous given where you were. –  Ross Millikan May 16 '11 at 13:43
    
@ Ross,please have a look on my logic just below the question –  prem shekhar May 16 '11 at 13:54

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